CODEWITHSUNDEEP
ccompiler-errors
xKalelX
xKalelX

Reputation: 99

Why am I getting this error in C? incompatible types

15:9: error: incompatible types when assigning to type ‘char[3]’ from type ‘char *’

#include <stdio.h>


int main(int argc, char *argv[])
{

     char servIP[3];
     int servPortNum;
     if(argc<3)
     {
         printf("Usage: clientApp servIP servPortNum\n");
     }

     servIP = argv[1];
     servPortNum = atoi(*argv[2]);


}

Upvotes: 0

Views: 1185

Answers (4)

asaelr
asaelr

Reputation: 5456

You can't assign an array like this. Assign it member-by-member, or use char *servIP instead.

Upvotes: 0

ouah
ouah

Reputation: 145919

You cannot assign to arrays. Use strcpy or strncpy function to copy a string in an array of char.

Upvotes: 1

smparkes
smparkes

Reputation: 14083

servIP is an array, not a pointer. Arrays convert to pointers, but they aren't the same thing and pointers don't convert to arrays.

Upvotes: 0

James M
James M

Reputation: 16728

strncpy (servIP, argv [1], sizeof (servIP) - 1);
servIP [sizeof (servIP) - 1] = 0;

But are you sure servIP is big enough for an IP address?

Upvotes: 2

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